Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $k = \dfrac{-4x + 12}{x^2 - 11x + 24} \div \dfrac{3x + 18}{-x^2 + 10x - 16} $
Dividing by an expression is the same as multiplying by its inverse. $k = \dfrac{-4x + 12}{x^2 - 11x + 24} \times \dfrac{-x^2 + 10x - 16}{3x + 18} $ First factor out any common factors. $k = \dfrac{-4(x - 3)}{x^2 - 11x + 24} \times \dfrac{-(x^2 - 10x + 16)}{3(x + 6)} $ Then factor the quadratic expressions. $k = \dfrac {-4(x - 3)} {(x - 8)(x - 3)} \times \dfrac {-(x - 8)(x - 2)} {3(x + 6)} $ Then multiply the two numerators and multiply the two denominators. $k = \dfrac {-4(x - 3) \times -(x - 8)(x - 2) } { (x - 8)(x - 3) \times 3(x + 6)} $ $k = \dfrac {4(x - 8)(x - 2)(x - 3)} {3(x - 8)(x - 3)(x + 6)} $ Notice that $(x - 8)$ and $(x - 3)$ appear in both the numerator and denominator so we can cancel them. $k = \dfrac {4\cancel{(x - 8)}(x - 2)(x - 3)} {3\cancel{(x - 8)}(x - 3)(x + 6)} $ We are dividing by $x - 8$ , so $x - 8 \neq 0$ Therefore, $x \neq 8$ $k = \dfrac {4\cancel{(x - 8)}(x - 2)\cancel{(x - 3)}} {3\cancel{(x - 8)}\cancel{(x - 3)}(x + 6)} $ We are dividing by $x - 3$ , so $x - 3 \neq 0$ Therefore, $x \neq 3$ $k = \dfrac {4(x - 2)} {3(x + 6)} $ $ k = \dfrac{4(x - 2)}{3(x + 6)}; x \neq 8; x \neq 3 $